A beam of light has three wavelengths `4144 Å`, `4972 Å` and `6216 Å` with a total instensity of `3.6 xx 10^(-3) Wm^(-2)` equally distributed amongst the three wavelengths. The beam falls normally on an area `1.0 cm^2` of a clean metallic surface of work function 2.3 eV. Assume that there is no loss of light by reflection and that each energetically capable photon ejects on electron. Calculate the number of photo electrons liberated in two seconds.

Explanation

Three different wavelengths are incident on metal surface, so first determine which is (are ) capable of ejecting photo- electrons.
For photo - emission , `lamda le lamda_(0) ` Given : `phi_0 = 2.3 eV`
`phi _(0) = hc//lamda_(0)`
` implies lamda_(0) =(hc)/(phi_0) = (6.63 xx10^(-34) xx 3 xx 10^(8))/(2.3 xx 1.6 xx 10^(-19)) = 5404 Å`
`implies` only wavelengths `4144 Å and 4972 Å` will cause photo - emission `(6216Å gt lamda-0)`
Intensity of each incident wavelength
` 3.6 xx(10^(-3))/3 =1.2 xx 10^(-3) W//m^2`
[ `:.` I is distributed equally among three wavelengths ]
Number of electrons emitted per second , n/sec `= (IA)/(hc//lamda)`
n/sec `(lamda = 4144 Å)`
`= ((1.2 xx10^(-3))xx(10^(-4)) xx 4972xx10^(-10))/(6.63 xx 10 ^(-34)xx3xx10^8)=2.5xx10^(11)`
`n//sec(lamda = 4972 Å)`
`= ((1.2 xx10^(-3))xx(10^(-4)) xx 4972xx10^(-10))/(6.63 xx 10 ^(-34)xx3xx10^8)=3xx10^(11)`
`implies` total electrons emitted /sec = `5.5 xx 10^(11) `

`implies` total electrons emitted in 2 seconds `11 xx 10^(11)`
`eVs =1 /2 mv_(max)^2`